Elementary Linear Algebra 7th Edition

Published by Cengage Learning
ISBN 10: 1-13311-087-8
ISBN 13: 978-1-13311-087-3

Chapter 3 - Determinants - Review Exercises - Page 138: 29


$|A|=6$ and $|A^{-1}|==\frac{1}{6}$.

Work Step by Step

Let $A$ be given by $$A=\left[ \begin {array}{cccc} 1&-1\\ 2&4 \end {array} \right] ,$$ then we have $A^{-1}=\frac{1}{6}\left[ \begin {array}{cccc} 4&1\\ -2&1 \end {array} \right]$. Now, $|A|=6$ and $|A^{-1}|=\frac{1}{6^2}6=\frac{1}{6}$. Hence, we verify that $$|A^{-1}=\frac{1}{|A|}.$$
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